3D-kreativ - 10 months ago 35

Android Question

There's a lot of questions about intersections between line segments here at stackowerflow and here is one more! Sorry, but I need help to understand how to calculate intersections. I have read several of the questions here and looked at several examples on other websites, but I'm still confused and don't get it! I don't like to copy and paste code without how things work.

So far I know that I'm going to compare the points of each line segments like Ax, Ay, Bx, By, Cx, Cy, Dx, Dy. Could someone explain for me what I'm going to calculate, what would the result of the calculating be if there is an intersection?

This is one of the example code I seen. I guess I don't need the intersecting point, just to know if the lines intersect or not.

`public static Point lineIntersect(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) {`

double denom = (y4 - y3) * (x2 - x1) - (x4 - x3) * (y2 - y1);

if (denom == 0.0) { // Lines are parallel.

return null;

}

double ua = ((x4 - x3) * (y1 - y3) - (y4 - y3) * (x1 - x3))/denom;

double ub = ((x2 - x1) * (y1 - y3) - (y2 - y1) * (x1 - x3))/denom;

if (ua >= 0.0f && ua <= 1.0f && ub >= 0.0f && ub <= 1.0f) {

// Get the intersection point.

return new Point((int) (x1 + ua*(x2 - x1)), (int) (y1 + ua*(y2 - y1)));

}

return null;

}

Do I also need to calculate some median value like in this code example?

`For lines through points (x0,y0) and (x1,y1), let xm = (x0+x1)/2, ym = (y0+y1)/2 (median of line segment).`

Then a = (y1-y0) and b = (x0-x1).

If you evaluate c = a(x-xm)+b(y-ym), c=0 for (x,y) on the line, and the sign(c) tells you which side a point is on

Answer Source

The first piece of code you show is based on vector cross-product, which has been explained here How do you detect where two line segments intersect? in a great detail.

IMO, an easier way of understanding it is through solving a system of equations. Firstly look at lines generally and then cut segments from them. Below I use notations for given segments `((x1, x2), (y1, y2))`

and `((x3, x4), (y3, y4))`

.

Check if any of the lines is vertical (

`x1 == x2`

or`x3 == x4`

).a. If both are vertical and

`x1 != x3`

, then no intersection.b. If both are vertical and

`x1 == x3`

, check if`(y1, y2)`

and`(y3, y4)`

overlap.c. If only one is vertical (say, first one), then build the equation of the second line (as outlined below), find the point where two lines intersect (by substituting

`x1`

into the equation of the second line) and check if this point is within both segments (similar to step 5).d. If not, proceed.

Use the points coordinates to build lines equations in form

`y = a*x + b`

(like here).`a1 = (y2-y1)/(x2-x1) b1 = y1 - a1*x1 a2 = (y4-y3)/(x4-x3) b2 = y3 - a2*x3`

Check if lines are parallel (same slope

`a`

). If yes, check if they have same intercept`b`

. If yes, check if 1D segments`(x1, x2)`

and`(x3, x4)`

overlap. If yes, your segments do overlap. The case when lines are parallel can be ambiguous. If they overlap, you can consider it as an intersection (it even can be one point if their ends are touching), or not. Note: if you are working with floats it would be a bit trickier, I reckon you would want to ignore this. In case you have only integers checking if`a1 = a2`

is equivalent to:`if((y2-y1)*(x4-x3) == (x2-x1)*(y4-y3))`

If lines are not parallel. The point of intersection is equivalent to a solution of a system of equations representing the two lines. Really,

`y = a1*x + b1`

and`y = a2*x + b2`

intersecting basically means that both of these equations hold. Solve this system by equating the two right sides and it will give you the intersection point. In fact, you need only`x`

coordinate of the intersection (draw it and you'll see why):`x0 = -(b1-b2)/(a1-a2)`

Last step is to check if the intersection point

`x0`

lies within both segments. That is,`min(x1, x2) < x0 < max(x1, x2)`

and`min(x3, x4) < x0 < max(x3, x4)`

. If yes, your lines do intersect!